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The magnitude of average acceleration in half-time period from equilibrium position in a simple harmonic motion is,

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Important Questions on Simple Harmonic Motion

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Alpha Question Bank for Engineering: Physics>Chapter 19 - Simple Harmonic Motion>Exercise - 1>Q 34
A particle performs SHM on the y-axis according to equation, y=A+Bsinωt. Its amplitude is,
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Alpha Question Bank for Engineering: Physics>Chapter 19 - Simple Harmonic Motion>Exercise - 1>Q 35
Two particles execute S.H.M. of same amplitude and frequency along the same straight line from same mean position. They cross one another without collision when going in opposite directions. Each time, their displacement from mean position is half of their amplitudes. The phase difference between them is,
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Alpha Question Bank for Engineering: Physics>Chapter 19 - Simple Harmonic Motion>Exercise - 1>Q 36
A mass M performs linear simple harmonic motion. Then, the correct graph for acceleration a and corresponding linear velocity v is,
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Alpha Question Bank for Engineering: Physics>Chapter 19 - Simple Harmonic Motion>Exercise - 1>Q 37
A body executing SHM passes through its equilibrium. At this instant, it has,
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Alpha Question Bank for Engineering: Physics>Chapter 19 - Simple Harmonic Motion>Exercise - 1>Q 38
The K.E. and P.E of a particle executing SHM with amplitude A will be equal when its displacement is
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Alpha Question Bank for Engineering: Physics>Chapter 19 - Simple Harmonic Motion>Exercise - 1>Q 39
A point particle of mass 0.1 kg executes S.H.M. of amplitude of 0.1 m. When the particle passes through the mean position, its kinetic energy is 8 × 10–3 J. The equation of motion of this particle when the initial phase of oscillation is 45° can be given by,
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Alpha Question Bank for Engineering: Physics>Chapter 19 - Simple Harmonic Motion>Exercise - 1>Q 40
For a particle performing SHM,
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Alpha Question Bank for Engineering: Physics>Chapter 19 - Simple Harmonic Motion>Exercise - 1>Q 41

Acceleration a versus time t graph of a body in SHM is given by a curve shown below. T is the time period. Then, the corresponding graph between kinetic energy KE and time t is correctly represented by,

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